A person, standing exactly midway between two towers, observes the top of the two towers at angle of elevation of 22.5° and 67.5°. What is the ratio of the height of the taller tower to the height of the shorter tower? (Given that tan 22.5° = √2−1)
1−2√2 : 1
1+2√2 : 1
3+2√2 : 1
3−2√2 : 1
Explanation:
Let ED be the taller tower and AB be the shorter tower.
Let C be the point of observation.
Given that $\angle$ACB = 22.5° and $\angle$DCE = 67.5°
Given that C is the midpoint of BD.
Hence, BC = CD
From the right $\triangle$ ABC,
tan22.5°=$\dfrac{AB}{BC}$ ⋯(eq:1)
From the right $\triangle$ CDE,
tan67.5°=$\dfrac{ED}{CD} $ ⋯(eq:2)
$\dfrac{(eq:2)}{(eq:1)}⇒\dfrac{tan67.5°}{tan22.5°} =\dfrac{\left( \dfrac{ED}{CD} \right)}{\left( \dfrac{AB}{BC} \right)}$
=$\dfrac{ED}{AB}$(∵ CD=BC)
⇒$\dfrac{tan(90°−22.5°)}{tan22.5°}=\dfrac{ED}{AB}$
⇒$\dfrac{cot22.5°}{tan22.5°}=\dfrac{ED}{AB}$ [∵ tan(90-θ)=cot θ]
⇒$\dfrac{\left( \dfrac{1}{tan22.5°}\right)}{tan22.5°}=\dfrac{ED}{AB }$ [∵cotθ=$\dfrac{1}{tanθ}$]
⇒$\dfrac{ED}{AB}=\dfrac{1}{(tan22.5°)^{2}}$
=$\dfrac{1}{(\sqrt{2}−1)^{2}}=\left( \dfrac{1}{\sqrt{2}−1} \right)^{2}$
=$\left[\dfrac{(\sqrt{2}+1)}{(\sqrt{2}−1)(\sqrt{2}+1)}\right]^{2}$
=$\left[\dfrac{(\sqrt{2}+1)}{(2−1)}\right]^{2}=\left[\dfrac{(\sqrt{2}+1)}{1}\right]^{2}$
=(√2+1)2=(2+2√2+1)
=(3+2√2)
Required ratio = ED : AB =(3+2√2):1
Let ED be the taller tower and AB be the shorter tower.
Let C be the point of observation.
Given that $\angle$ACB = 22.5° and $\angle$DCE = 67.5°
Given that C is the midpoint of BD.
Hence, BC = CD
From the right $\triangle$ ABC,
tan22.5°=$\dfrac{AB}{BC}$ ⋯(eq:1)
From the right $\triangle$ CDE,
tan67.5°=$\dfrac{ED}{CD} $ ⋯(eq:2)
$\dfrac{(eq:2)}{(eq:1)}⇒\dfrac{tan67.5°}{tan22.5°} =\dfrac{\left( \dfrac{ED}{CD} \right)}{\left( \dfrac{AB}{BC} \right)}$
=$\dfrac{ED}{AB}$(∵ CD=BC)
⇒$\dfrac{tan(90°−22.5°)}{tan22.5°}=\dfrac{ED}{AB}$
⇒$\dfrac{cot22.5°}{tan22.5°}=\dfrac{ED}{AB}$ [∵ tan(90-θ)=cot θ]
⇒$\dfrac{\left( \dfrac{1}{tan22.5°}\right)}{tan22.5°}=\dfrac{ED}{AB }$ [∵cotθ=$\dfrac{1}{tanθ}$]
⇒$\dfrac{ED}{AB}=\dfrac{1}{(tan22.5°)^{2}}$
=$\dfrac{1}{(\sqrt{2}−1)^{2}}=\left( \dfrac{1}{\sqrt{2}−1} \right)^{2}$
=$\left[\dfrac{(\sqrt{2}+1)}{(\sqrt{2}−1)(\sqrt{2}+1)}\right]^{2}$
=$\left[\dfrac{(\sqrt{2}+1)}{(2−1)}\right]^{2}=\left[\dfrac{(\sqrt{2}+1)}{1}\right]^{2}$
=(√2+1)2=(2+2√2+1)
=(3+2√2)
Required ratio = ED : AB =(3+2√2):1